[数论+二分求等比数列]POJ 1845 Sumdiv

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题目链接：http://poj.org/problem?id=1845

题目大意：求A^B 所有因子的和% 9901

思路：A = p1^a1 * p2^a2 * ... ... * pn^a3 那么A的所有因子之和为 sigma ( pi^0 + pi^1 + pi^2 + ... ... + pi^ai )

A^B 的所有因子之和为 sigma ( pi^0 + pi^1 + pi^2 + ... ... + pi^（ai*B ))

对于 ( pi^0 + pi^1 + pi^2 + ... ... + pi^（ai*B )) 这是一个等比数列，可以用二分求解，时间复杂度为logN

Code：

#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>#include <cmath>using namespace std;#define foru(i, a, b) for (int i=a; i<=b; i++)#define ford(i, a, b) for (int i=a; i>=b; i--)#define ll __int64#define M 9901#define N 10001bool f[N];int pri[N], m = 0 ;void getpri(){    foru(i, 2, N-1){        if (! f[i]) {m++; pri[m] = i;}        foru(j, 1, m){            if (i * pri[j] > N-1) break;            f[i * pri[j]] = 1;            if (i % pri[j] == 0) break;        }    }}ll g(ll a, ll b){    ll s = 1;    while (b){        if ((b&1) == 1) s = (s*a)%M;        b >>= 1;        a = (a*a)%M;    }    return s;}ll x, y;ll getsum(ll p, ll n){    if (! n) return 1ll;    if (n&1) return (getsum(p, n/2) * (1 + g(p, n/2+1)))%M;    return (getsum(p, n/2-1) * (1 + g(p, n/2+1)) + g(p, n/2))%M;}ll get(){    ll sum = 1ll;    foru(i, 1, m){        if (x == 1) break;        ll t = 0ll;        while (x%pri[i] == 0){            t ++;            x /= pri[i];            if (x == 1) break;        }        if (t == 0) continue;        sum = (sum * getsum(pri[i], t*y))%M;    }    if (x > 1) sum = (sum * getsum(x, y))%M;    while (sum < 0) sum += M;    return sum;}int main(){    //freopen("K.txt", "r", stdin);    getpri();    while (scanf("%I64d %I64d", &x, &y)!= EOF ){        if (! x && ! y) printf("%I64d\n", 1ll);            else if (! x) printf("%I64d\n", 0ll);                else if (! y)printf("%I64d\n", 1ll);                    else printf("%I64d\n", get());    }    return 0;}

Tips：

二分求等比数列pi^0 + pi^1 + pi^2 + ... ... + pi^n

1°若n是偶数，那么等比数列一共有奇数项，（1 + pi^1 + pi^2 + ... ... + pi^(n/2-1）) * (1 +pi^(n/2+1） ) + pi^(n/2）最高次 (n/2-1) +（n/2+1) = n, 加上中间漏掉的n/2项

2°若n是奇数，那么等比数列一共有偶数项，（1 + pi^1 + pi^2 + ... ... + pi^(n/2）) * (1 +pi^(n/2+1） ) 最高次 (n/2） +（n/2+1) = n

ll getsum(ll p, ll n){    if (! n) return 1ll;    if (n&1) return (getsum(p, n/2) * (1 + g(p, n/2+1)))%M;    return (getsum(p, n/2-1) * (1 + g(p, n/2+1)) + g(p, n/2))%M;}

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